The COD value of a water sample is 40 ppm. Calculate the amount of acidified `K_(2)Cr_(2)O_(7)` required to oxidise the organic matter present in 500 ml of that water sample. 

Explanation

COD value is 40 ppm. It means 106g of water sample require 40 g of oxygen to oxidise the organic matter in it.
` 500g " water " to (40 xx 500)/(10^6)=2 xx 10^(-2) g " of " O_2`
500 mL water sample requires ` 2 xx 10^(-2) g ` of `O_2` , to oxidise the organic matter present in it .
`8g O_2 -= 49 g K_2Cr_2O_7`
` 2 xx 10^(-2) g " of " O_2 -= (49 xx 2 xx 10^-2)/(8) g " of " K_2Cr_2O_7`
Amount of `K_2Cr_2O_7` required to oxidise the organic matter present in the water sample is 0.1225 g .